Using vectors, prove that the perpendicular bisectors of the sid

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsShort Answer Type

221.
Prove that an angle inscribed in a semi-circle is a right angle.
87 Views

 Multiple Choice QuestionsLong Answer Type

222. Using vectors, prove that if two medians of a triangle ABC be equal, then it is an isosceles triangle. 
157 Views

Advertisement

223. Using vectors, prove that the perpendicular bisectors of the sides of a triangle are concurrent.


Let ABC be the given triangle and O be the point of intersection of perpendicular bisectors OD and OE of sides BC and CA respectively.
Let F be mid-point of AB. Join O to F.
Take O as origin.

Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top be the position vectors of A, B, C respectively.
            therefore space space OA space equals space straight a with rightwards arrow on top comma space space OB with rightwards arrow on top space equals space straight b with rightwards arrow on top comma space space OC with rightwards arrow on top space equals straight c with rightwards arrow on top
The position vectors of D, E, F are fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction comma space space fraction numerator straight c with rightwards arrow on top plus straight a with rightwards arrow on top over denominator 2 end fraction comma space fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator 2 end fraction respectively.
Since               OD space perpendicular space BC
therefore space space space space space space space OD with rightwards arrow on top. space BC with rightwards arrow on top space equals space 0
rightwards double arrow space space space space open parentheses fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction close parentheses. space left parenthesis straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space space 0 space space space space rightwards double arrow space space 1 half left parenthesis straight c with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space left parenthesis straight c with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space space space 1 half left parenthesis straight c with rightwards arrow on top squared minus straight b with rightwards arrow on top squared right parenthesis space equals space 0
Again, OE space perpendicular space CA
therefore space space space OE with rightwards arrow on top. space CA with rightwards arrow on top space equals space 0 space space space space rightwards double arrow space space space open parentheses fraction numerator straight c with rightwards arrow on top plus straight a with rightwards arrow on top over denominator 2 end fraction close parentheses. space left parenthesis straight a with rightwards arrow on top. space straight c with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space space space 1 half left parenthesis straight a with rightwards arrow on top squared minus straight c with rightwards arrow on top squared right parenthesis space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Adding (1) and (2), we get,
              1 half left parenthesis straight a with rightwards arrow on top space minus straight b with rightwards arrow on top squared right parenthesis space space space space rightwards double arrow space space space 1 half left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space space space open parentheses fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator 2 end fraction close parentheses space. space left parenthesis straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis space equals space stack OF. with rightwards arrow on top space AB with rightwards arrow on top space equals space 0
rightwards double arrow space space space space OF space perpendicular space AB
therefore    perpendicular bisectors meet in a point. 

274 Views

Advertisement

 Multiple Choice QuestionsShort Answer Type

224. Prove that the median to the base of isosceles triangle is perpendicular to the base.
257 Views

Advertisement

 Multiple Choice QuestionsLong Answer Type

225. If the median of the base of a triangle is perpendicular on the base, then prove that the triangle is an isosceles.
90 Views

226. (i) Prove that cos (α + β) )= cos α cos β – sin α sin β.
(ii)  Prove that cos (α – p) = cos α cos β + sin α sin β.
85 Views

227. Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
87 Views

 Multiple Choice QuestionsShort Answer Type

228. Using vectors, prove that a parallelogram whose diagonals are equal is a rectangle. 
88 Views

Advertisement

 Multiple Choice QuestionsLong Answer Type

229. Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.
78 Views

230. Show that the angle between two diagonals of a cube is cos to the power of negative 1 end exponent space open parentheses 1 third close parentheses.
82 Views

Advertisement